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April 19, 2014

April 19, 2014

Posted by **anoynomous** on Friday, March 15, 2013 at 1:55pm.

area of parallelogram APQR=( 1/2)* area of triangle[ABC]

- maths-areas -
**Reiny**, Friday, March 15, 2013 at 2:09pmJoin PR

By the mid-point of triangle theorem,

PR is || to BC and PR = (1/2)BC

Also triange APR is similar to triange ABC

so the areas are proportional to the square of their sides.

since the sides are 1:2

their areas are 1:4

so triangle APR = (1/4) of triangle ABC

Similary BQP would be 1/4 of triangle ABC, and

RQC is 1/4 of triangle ABC, leaving the inside triangle PQR also as 1/4 of triangle ABC to get 4/4

so figure APQR = 2/4 or 1/2 of triangle ABC

- maths-areas -
**anoynomous**, Friday, March 15, 2013 at 2:22pmHow traingle ABC and APR are similar

- maths-areas -
**Reiny**, Friday, March 15, 2013 at 2:32pmSince BC and PR are similar

angle B = angle APR

and angle A is common

if 2 angles of a triangle are equal to 2 corresponding angles of another triangle, the triangles are similar.

Same argument for the other pairs of similar triangles.

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